Question

Fred wants to sell his car, after moving back to Blissville (where he is happy with the bus system). He decides to sell it to the first person to offer at least $18,000 for it. Assume that the offers are independent Exponential random variables with mean $12,000, and that Fred is able to keep getting offers until he obtains one that meets his criterion.

Find the expected number of offers Fred will have.

Answer 1

Expect ≈1.58 offers, as Fred waits for a buyer to meet his $18,000 minimum.

In this scenario, Fred is essentially waiting for the first "success" event, which is an offer of at least $18,000. The arrival of such offers follows a geometric distribution with parameter p, the probability of receiving an acceptable offer.

Here's how to find the expected number of offers:

1. Probability of an acceptable offer:

The mean of the exponential distribution is $12,000.

We want the probability of an offer to be less than $18,000 (since Fred wants more).

Using the exponential distribution formula, p(offer < $18,000) = 1 - exp(-18,000 / 12,000) ≈ 0.368.

2. Probability of needing multiple offers:

This is simply 1 - p, the probability of not getting an acceptable offer on the first try.

So, 1 - 0.368 ≈ 0.632.

3. Expected number of offers:

This is the average number of tries it takes to get a success in a geometric distribution.

It's calculated as 1 / (probability of needing multiple offers).

Therefore, the expected number of offers for Fred is 1 / 0.632 ≈ 1.58.

Therefore, Fred can expect to receive approximately 1.58 offers before someone offers at least $18,000 for his car.